/sci/ - Science & Math

[Anonymous No.4025430]

/sci/ i need your help with optimization problems

A cylindrical can is to have volume 1900 cubic centimeters. Determine the radius and the height which will minimize the amount of material to be used.
Note that the surface area of a closed cylinder is S=2πrh+2πr2 and the volume of a cylindrical can is V=πr2h.


My work:
1900 = πr^2h
h = 1900/(πr^2)

S = 2πrh + 2πr^2
S = 2πr(1900/(πr^2)) + 2πr^2
S = 3800/r + 2πr^2
S' = -3800/r^2 + 4πr
0 = -3800/r^2 + 4πr
0 = r(4π - 3800/r^2)
0 = r //r > 0

0 = 4π - 3800/r^2
-3800/r^2 = -4π
3800/4π = r^2

r = sqrt(3800/(4π))

h = 1900/(π(sqrt(3800/(4π))^2))
h = 1900/(π(3800/(4π)))
Apparently I'm wrong and i don't know why.

[Anonymous No.4025448]

tau > pi

[Anonymous No.4025456]

>>4025448
It's meant to be a pi, not tau. Still doesn't change my problem here.

[Anonymous No.4025457]

Write out a function for area, using the formula for volume find either h as a function of r, or r as a function of h.
Plug that expression in the function for area so that you have area as a function of either h or r.
Differentiate, find the critical values.
Depending on which variable you used, plug your result in the formula for volume to find the second one.

[Anonymous No.4025468]

>>4025457
That's exactly what i did.

V = volume function
S = surface area function

found h in terms of r, sub'd into S
Differentiated S, found criticals plugged in and it was wrong

[Anonymous No.4025479]

0 = -3800/r^2 + 4πr

3800 = 4pi(r)^3

[Anonymous No.4025486]

>>4025430
This step is wrong.

>>0 = -3800/r^2 + 4πr
>>0 = r(4π - 3800/r^2)

There will be an r^3 somewhere.

[Anonymous No.4025521]

>>4025479
>>4025486
oh man i derp'd that's the correct answer with cube root instead. thanks


I have another problem :

An open box is to be constructed so that the length of the base is 5 times larger than the width of the base. If the cost to construct the base is 5 dollars per square foot and the cost to construct the four sides is 1 dollars per square foot, determine the dimensions for a box to have volume = 85 cubic feet which would minimize the cost of construction.

I am to find the dimensions.


My Work:

L = 5w
85 = 5w^2h
h = 85/(5w^2)


sa = 5w^2 + 2(wh) + 2(5wh)
sa(w) = 5w^2 + 12(85/(5w^2))w
sa(w) = 5w^2 + 204/w
sa'(w) = 10w - 204/w^2

0 = 10w - 204/w^2
0 = w(10 - 204/w^3)
w = 0 // w > 0 domain

w^3 = 204/10
w = (204/10)^(1/3)

L = 5 * (204/10)^(1/3)

h = 85/(5 * (204/10)^(1/3))


this is apparently wrong to and it's probably because i didn't incorporate the cost per square foot for base/sides, but i'm not sure how i would incorporate. If anyone could hint me at it i would love you.

[Anonymous No.4025557]

>>4025521
>>sa = 5w^2 + 2(wh) + 2(5wh)

cost = 5*5w^2 + 2(wh) + 2(5wh)

[Anonymous No.4025574]

>>4025557
K got that one now, thank you.


Last question:
Using Newton's Method, determine the first approximation to the solution of:
e^(3.5x)=6−5x

starting with x = 0.

This one should be easy but for some reason i'm doing it wrong.

f(x) = 6-5x-e^(3.5x)
f ' (x) = -5-3.5e^(3.5x)

x = 0;

x1 = x - f(x)/f ' (x)

x1 = 0 - (6-0-e^0)/(-5-3.5e^0)
x1 = (6-1)/(-5-3.5)
x1 = (5)/(-8.5)

this is also wrong apparently.

Did i do something wrong?